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3 years ago

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A hot iron ball is dropped into 200.0 g of cool water. The water temperature increases by 2.0 C and the temperature of the ball

decreases by 18.6 C. What is the mass of the ball? waters specific heat is 4184 (J/kg•C)

1 answer:

Bess [88]3 years ago

7 0

0.2 KG, just search the question and it's on yahoo answers

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A measure of randomness or disorder

vodka [1.7K]

It is simply called Entropy.

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3 years ago

Read 2 more answers

How do the principles of convection, conduction, and radiation explain how the water in the saucepan gets hot?

RSB [31]

<span>Heat comes from stove flame to the sauce pan by radiation through infrared energy, heat conducts the metal of the sauce pan; Convection brings cool water to the hot surface at the bottom of the hot sauce pan until all or most of the water is hot enough to boil.</span>

3 0

3 years ago

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.

VMariaS [17]

Answer:

A) $d=\dfrac{1}{2F}mv^2$

B) $\Delta KE'=2\times \dfrac{1}{2}mv^2$

Explanation:

Given that

Force = F

Increase in Kinetic energy = $\dfrac{1}{2}mv^2$

$\Delta KE=\dfrac{1}{2}mv^2$

we know that

Work done by all the forces =change in the kinetic energy

a)

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

$F.d=\dfrac{1}{2}mv^2$

$d=\dfrac{1}{2F}mv^2$

b)

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE' ( F.d =Δ KE)

2ΔKE = ΔKE'

$\Delta KE'=2\times \dfrac{1}{2}mv^2$

Therefore the final kinetic energy will become the twice if the force become twice.

8 0

4 years ago

tyhe mass of an object is 117 g adding 1200j of heat will raise the temperture of the object by 12 celsius what is the specifc h

Sati [7]

Answer:

C 0.85 j/g*k

Explanation:

The specific heat capacity of a material is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q is the amount of heat supplied to the object

m is the mass of the object

$\Delta T$ is the increase in temperature of the object

For the object in this problem, we have

m = 117 g is the mass

Q = 1200 J is the heat supplied

$\Delta T=12^{\circ}$ is the increase in temperature

Substituting into the formula, we find the specific heat:

$C_s = \frac{1200 J}{(117 g)(12^{\circ})}=0.85 J/gC$

8 0

4 years ago

A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hang

valina [46]

Answer:

work done is -2.8 × 10⁻⁶ J

Explanation:

Given the data in the question;

mass of the pendulum m = 6 kg

Length of core = 1.7 m

Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad

In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.

so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.

Now, the required work done will be;

$W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta$

$W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta$

$W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta$

W = $-mgl[$ -cosθ $]^{0.047}_{0.045 }$

W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]

W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]

W = -2.8 × 10⁻⁶ J

Therefore, work done is -2.8 × 10⁻⁶ J

6 0

3 years ago